Ponctual electric charge problem

[fishingspree2]

Hello everyone, my answer on the following problem is wrong and I don't see why. I am translating the question from another language, I am sorry if the english is not clear.

Let q₁ = 27C localised at x=0, and another charge q₂=3C localised at x=1 meters.

a)Besides infinity, find a point where the net force exerced on a third ponctual charge would be = 0.

Solution:
The third charge must be located between q₁ and q₂ so the forces exerted cancel out.
If is the distance between q₁ and the third charge, then 1-d is the distance between q₂ and the third charge. Let q_x be the third charge.

Then by Coulomb's law:
$$0=\frac{kq_{x}q_{1}}{d^{2}}+\frac{kq_{x}q_{2}}{1-d^{2}}$$
k and q_x cancel out, we now have
$$0=27-54d+30d^{2}$$
This equation has no solution.

However, if we look at things intuitevely, there must be a point where the forces cancel out. the answer is x=0,750 meters.
What is wrong?
Sorry for my english

 

[Doc Al]

Then by Coulomb's law:
$$0=\frac{kq_{x}q_{1}}{d^{2}}+\frac{kq_{x}q_{2}}{1-d^{2}}$$

Two errors here:
(1) The denominator in the second term should be (1-d)^2, not 1-d^2.
(2) The sign of the second term must be negative, since the field from q2 points to the left.

 

[cepheid]

From your Coulomb equation I get d^2(q2) = -(1 - d^2)q1 = q1d^2 - q1(1m)

(q2 - q1)d^2 = -q1(1 m)

-24d^2 = -27(1 m)

d^2 = 27/24 = 9/8

d = 3/2*sqrt(2)

It doesn't match your answer, so maybe I messed up. HOWEVER the point of my post is that, your quadratic equation is clearly wrong. Where did you get the d term from?

EDIT: Okay, so DocAl pointed out where the mistake was...